
What is the gauge pressure at the water mercury interface?
Answer
484.2k+ views
Hint
We know that the water mercury interface does not come under the influence of the atmospheric pressure. This makes us curtail the value of atmospheric pressure at any point of the calculation.
${P_g} = \rho gh$
Where, ${P_g}$ is the gauge pressure, $\rho $is the density of the material, $g$is the acceleration due to gravity and $h$ is the height or depth of the material from the interface.
Complete step by step answer
We know, Density of water, ${\rho _w} = {10^3}kg{m^{ - 3}}$
Height of water from the interface, ${h_w} = 15cm = 0.15m$
Acceleration due to gravity, $g = 9.81m{s^{ - 2}}$
Now, Substituting these value in the formula, we get,
$\Rightarrow {P_g} = {10^3} \times 9.81 \times 0.15N{m^{ - 2}}$
By evaluating and converting to scientific notation, we get
$\Rightarrow {P_g} = 1.47 \times {10^3}Pa$
Hence, the gauge pressure at the water mercury interface is $1.47 \times {10^3}Pa$.
Additional Information
Pressure measurement is known to be the analysis of an applied force by a fluid on a surface. This is typically measured in units of force per unit of surface area. Up till now, many practices or techniques have been developed for the measurement of pressure and vacuum. Gauge Pressure refers to atmospheric pressure. It is found to be positive for pressures above atmospheric pressure, and negative for pressures below it. In this case there was no scope for atmospheric pressure to act. But if anyhow, the atmospheric pressure comes into play, the total pressure will be different from the gauge one as the atmospheric pressure will be added to this and that will make the change.
Note
The values are to an extreme precision. So be careful while calculating. The higher precision calculations might get clumsier.
We know that the water mercury interface does not come under the influence of the atmospheric pressure. This makes us curtail the value of atmospheric pressure at any point of the calculation.
${P_g} = \rho gh$
Where, ${P_g}$ is the gauge pressure, $\rho $is the density of the material, $g$is the acceleration due to gravity and $h$ is the height or depth of the material from the interface.
Complete step by step answer
We know, Density of water, ${\rho _w} = {10^3}kg{m^{ - 3}}$
Height of water from the interface, ${h_w} = 15cm = 0.15m$
Acceleration due to gravity, $g = 9.81m{s^{ - 2}}$
Now, Substituting these value in the formula, we get,
$\Rightarrow {P_g} = {10^3} \times 9.81 \times 0.15N{m^{ - 2}}$
By evaluating and converting to scientific notation, we get
$\Rightarrow {P_g} = 1.47 \times {10^3}Pa$
Hence, the gauge pressure at the water mercury interface is $1.47 \times {10^3}Pa$.
Additional Information
Pressure measurement is known to be the analysis of an applied force by a fluid on a surface. This is typically measured in units of force per unit of surface area. Up till now, many practices or techniques have been developed for the measurement of pressure and vacuum. Gauge Pressure refers to atmospheric pressure. It is found to be positive for pressures above atmospheric pressure, and negative for pressures below it. In this case there was no scope for atmospheric pressure to act. But if anyhow, the atmospheric pressure comes into play, the total pressure will be different from the gauge one as the atmospheric pressure will be added to this and that will make the change.
Note
The values are to an extreme precision. So be careful while calculating. The higher precision calculations might get clumsier.
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